Find the values of p and q such that 4x^2+12x=4(x+p)^2-q

Two solutions were found :

x =(3-√13)/2=-0.303

x =(3+√13)/2= 3.303

Step-by-step explanation:

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "x2"   was replaced by   "x^2".

Step by step solution :

Step  1  :

Equation at the end of step  1  :

 (22x2 -  12x) -  4  = 0

Step  2  :

Step  3  :

Pulling out like terms :

3.1     Pull out like factors :

  4x2 - 12x - 4  =   4 • (x2 - 3x - 1)

Trying to factor by splitting the middle term

3.2     Factoring  x2 - 3x - 1

The first term is,  x2  its coefficient is  1 .

The middle term is,  -3x  its coefficient is  -3 .

The last term, "the constant", is  -1

Step-1 : Multiply the coefficient of the first term by the constant   1 • -1 = -1

Step-2 : Find two factors of  -1  whose sum equals the coefficient of the middle term, which is   -3 .

     -1    +    1    =    0

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Equation at the end of step  3  :

 4 • (x2 - 3x - 1)  = 0

Step  4  :

Equations which are never true :

4.1      Solve :    4   =  0

This equation has no solution.

A a non-zero constant never equals zero.

Parabola, Finding the Vertex :

4.2      Find the Vertex of   y = x2-3x-1

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) .   We know this even before plotting  "y"  because the coefficient of the first term, 1 , is positive (greater than zero).

Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.

Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.

For any parabola,Ax2+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is   1.5000  

Plugging into the parabola formula   1.5000  for  x  we can calculate the  y -coordinate :

 y = 1.0 * 1.50 * 1.50 - 3.0 * 1.50 - 1.0

or   y = -3.250

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = x2-3x-1

Axis of Symmetry (dashed)  {x}={ 1.50}

Vertex at  {x,y} = { 1.50,-3.25}

x -Intercepts (Roots) :

Root 1 at  {x,y} = {-0.30, 0.00}

Root 2 at  {x,y} = { 3.30, 0.00}

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step-by-step explanation:

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